Which of the following complexes will show geometrical isomerism?

First, recall that a coordination compound can exhibit geometrical isomerism only when it possesses at least two different kinds of ligands arranged round the central metal ion in such a way that two distinct spatial dispositions (for example, cis and trans) are possible. For an octahedral complex we need a situation like $$\text{[}MA_4B_2\text{]}$$ or $$\text{[}MA_2B_2C_2\text{]}$$, and for a square-planar complex we need something like $$\text{[}MA_2B_2\text{]}$$, $$\text{[}MA_2BC\text{]}$$, etc. If only one ligand is different, the complex usually cannot display geometrical isomerism because any possible arrangement can be rotated into another and becomes identical.

Now we examine each option in turn.

We have Option A: Potassium tris(oxalato)chromate(III), whose coordination entity is $$[Cr(C_2O_4)_3]^{3-}.$$ Each oxalato ligand $$C_2O_4^{2-}$$ is bidentate and all three are identical. Thus the entire octahedral sphere is occupied by three equivalent chelating ligands. Because no two different sets of ligands exist, only optical (Λ/Δ) isomerism is possible; geometrical isomerism is impossible. Hence Option A does not satisfy the required condition.

Next, Option B: Pentaaquachlorochromium(II) chloride, whose coordination entity is $$[Cr(H_2O)_5Cl]^+.$$ Here five ligands are $$H_2O$$ and only one ligand is $$Cl^-.$$ Writing the formula explicitly, we get $$\text{[}MA_5B\text{]}$$ with $$A = H_2O$$ and $$B = Cl^-.$$ Since there is only a single $$B$$ type ligand, rotating the octahedron makes every conceivable arrangement equivalent; thus no distinct cis or trans form can be created. So Option B also fails to show geometrical isomerism.

Consider Option C: Aquachlorobis(ethylenediamine)cobalt(II) chloride, whose coordination entity is $$[Co(en)_2(H_2O)Cl]^+. $$ Here $$en$$ (ethylenediamine) is a bidentate ligand and we have
$$2 \times en + 1 \times H_2O + 1 \times Cl^-,$$
giving four coordination sites from the two $$en$$ ligands and one site each from $$H_2O$$ and $$Cl^-,$$ for a total of six sites around the cobalt(II) ion, producing an octahedral geometry. Two different monodentate ligands, $$H_2O$$ and $$Cl^-,$$ occupy one position each. Now two distinct dispositions are possible:

• $$cis$$ form: the $$H_2O$$ ligand is adjacent to the $$Cl^-$$ ligand.
• $$trans$$ form: the $$H_2O$$ ligand is placed opposite to the $$Cl^-$$ ligand.

Because these two arrangements cannot be superimposed by simple rotation, real geometrical isomerism exists. Therefore Option C indeed shows geometrical (cis‒trans) isomerism.

Finally, Option D: Potassium ammine trichloroplatinate(II), whose coordination entity is $$[Pt(NH_3)Cl_3]^-. $$ Platinum(II) forms square-planar complexes. Writing the pattern explicitly we have $$\text{[}MA_3B\text{]}$$ with $$A = Cl^-$$ and $$B = NH_3.$$ In a square plane, every possible placement of the single $$NH_3$$ can be rotated to coincide with any other, so only one spatial arrangement exists; geometrical isomerism is therefore absent for Option D.

Summarising the above discussion, only Option C possesses the requisite two different monodentate ligands in an octahedral environment, leading to distinct cis and trans forms. None of the other options meet the geometric criterion.

Hence, the correct answer is Option C.